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\title{The Helmholtz-Weyl decomposition in $\Omega \subset \R^3$}
\author{Li Huiteng}

\begin{document}
\maketitle


This paper will show details of the Helmholtz-Weyl decomposition in an open bounded set in $\R^3$.
First we define the space of $H$ and $G$, then introduce Gauss Formula. Then we introduce
de Rham's Theorem and prove Helmholtz-Weyl decomposition.

\tableofcontents
\newpage
\section{Basic Concepts}
\subsection{Review and Notations}
\begin{defn}{\;}
    A function $u\in(L^1(\Omega))^3$ is weakly divergence free if
    $$
        \left \langle u,\nabla\phi  \right \rangle =0\text{ , }\forall \phi \in \cs(\Omega).
    $$
\end{defn}
\begin{nota}To avoid misunderstanding between spaces of scalar and vector-valued functions, we
    introduce the following notations.
    \begin{itemize}
        \item $\lbb^2:=[L^2]^3$
        \item $\hbb^s:=[H^s]^3$
        \item $\cdf(\Omega):=\{\phi \in [\cdf(\Omega)]^3:\diverg\phi=0\}$
    \end{itemize}
\end{nota}
\begin{defn}Let $\Omega$ be any open set of $R^3$ and the space $H=H(\Omega)$ is given by
    $$
        H\text{ := }\overline{\cdf(\Omega)}|_{\lbb^2(\Omega)}.
    $$
    Then $H$ is the completion of $\cdf$ in the norm of $\lbb^2$ and we equip the space $H$ with $\lbb^2$ norm.
\end{defn}
\begin{prop}\label{Hwdf}
    Any function in $H(\Omega)$ is weakly divergence free. Moreover, if $u \in H$ then
    $$
        \int_{\Omega}u(x)\cdot \nabla \Phi(x)dx = 0 \text{ , } \forall \Phi \in H^1(\Omega).
    $$
\end{prop}
\begin{proof}[\textbf{Proof.}]
    Let $u \in H,\Phi \in H^1(\Omega)$, and pick $\{\phi_n\} \in \cdf(\Omega)$ with $\phi_n\to u$ in $H$. Then $\forall n$,
    $$
        \int_{\Omega}\phi_n(x)\cdot \nabla \Phi(x)dx = -\int_{\Omega}(\diverg \phi_n)(x)\Phi(x)dx=0,
    $$
    where the first step follows from the definition of weak derivatives and the second step from $\diverg\phi_n=0$. Hence
    take $n \to \infty$ and we finish the proof.
\end{proof}
\subsection{Motivation of Trace}
Our goal is to show that any $u \in \lbb^2(\Omega)$ can be written in a unique way as the sum of
\begin{align}
    u=h+\nabla g,
\end{align}

where $h$ is divergence free and $g$ belongs to $H^1$. In other words
$$
    \lbb^2=H\oplus G,
$$
where $H$ is the space of divergence free functions and $G$ is the space of gradients of
functions in $H^1$.

When $\Omega$ is $\mathbb{R}^3$ or periodic domain,
the decomposition is easy to get by Fourier expansions(or transformations) and details can be found in Chapter 2.1 of the Book.

However, when $\Omega$ is a bounded open set, things turn different. A cruial matter to distinguish is that whether the completion of $\cdf(\Omega)$ is equivalent to the space of
divergence-free functions in $\lbb^2(\Omega)$. For $\mathbb{R}^3$ and $\mathbb{T}^3$, the statement is true.
The simplicity of decomposition is also because
boundaries are in absence in such two cases so that there's no need to worry about the trace.
But for open bounded domain, it doesn't hold since
boundaries should be taken into consideration. Although $\partial \Omega$ has n-dimensional Lebesgue measure zero, they can affect weak derivative and should be assigned carefully.

Therefore, we make the notion of trace operator and we then introduce several Trace Theorems which will be useful for the following chapters.
\begin{thm}{\bf{(Existence of Trace Operator)}}
    Assume $U$ is bounded and $\partial U$ is ${C}^1$. Then there exists a unique bounded linear operator
    $$
        T:W^{1,p}(U)\to L^p(\partial U),
    $$
    such that
    \begin{itemize}
        \item[(1)] $Tu=u|_{\partial U} \text{ if }u\in W^{1,p}(U)\cap C(\bar{U})$,
        \item[(2)] $\|Tu\|_{L^p(\partial U)}\leq C\|u\|_{W^{1,p}(U)}$, $\forall u \in W^{1,p}(U)$, and
              the constant $C$ only depends on $p$ and $U$.
    \end{itemize}
\end{thm}
\begin{defn}
    $Tu$ is called the trace of $u$ on $\partial U$.
\end{defn}
\begin{thm}\label{Kernel of Trace}{\bf{(Kernel of Trace Operator)}}
    Assume $U$ is bounded and $\partial U$ is ${C}^1$. Suppose $u \in W^{1,p}(U)$. Then
    $$
        u \in W^{1,p}_0(U)\text{ if and only if }Tu=0 \text{ on }\partial U.
    $$
\end{thm}
\begin{thm}\label{Lift of Trace}{\bf{(Image of Trace Operator and Existence of Lifting Operator)}}
    Suppose furthermore that $u \in H^1(\Omega)$ with trace operator $\gamma_0:H^1(\Omega)\to L^2(\partial\Omega)$. Then
    \begin{itemize}
        \item[(1)] $H^{\frac{1}{2}}(\partial \Omega)=\gamma_0(H^1(\Omega))$
              with $$\|v\|_{H^{\frac{1}{2}}(\partial \Omega)}=\inf_{u \in H^1(\Omega),\gamma_0(u)=v}\|u\|_{H^1(\Omega)}.$$
        \item[(2)] there exists a continous linear lifting operator $l_{\Omega}$, such that
              $$\gamma_0\cdot l_{\Omega}=\text{id. in }H^{\frac{1}{2}}(\partial\Omega).$$
    \end{itemize}
\end{thm}
\section{The Divergence Space and The Gradient Space}
To figure out the well-defineness of normal component of elements in $H(\Omega)$, we introduce the space $E(\Omega)$.
\subsection{The space $E(\Omega)$ and Gauss Formula}
\begin{defn}
    $E(\Omega)=\{u\in\lbb^2(\Omega),\diverg u \in L^2(\Omega)\}$,\par
    which is equipped with the norm $\|u\|_E=\left(\|u\|+\|\diverg u\|\right)^{1/2}$.
\end{defn}
\begin{thm}\label{Existence of normal component}
    {\bf{(Existence of normal component)}}Assume $\Omega$ is a bounded open set with smooth boundary.
    For $u \in E(\Omega)$, the normal component of $u$ is well defined on the boundary as a continous linear functional on the
    space of traces of functions in $H^1(\Omega)$, i.e. there exists
    $$\gamma_v:E\to H^{-1/2}(\partial\Omega),$$ such that
    \begin{itemize}
        \item[(1)] $\gamma_v u=(u\cdot n)|_{\partial \Omega} \text{ if }u\in (C_c^{\infty}(\overline{\Omega}))^3$,
        \item[(2)] For $u \in E(\Omega)$, denote $\gamma_u=\gamma_v(u)$. Then $$|\gamma_u\phi|\leq C\|\phi\|_{H^{1/2}(\partial\Omega)},$$ $\forall \phi \in H^{1/2}(\partial\Omega)$, and
              the constant $C$ only depends on $\Omega$ and $\|u\|_E$.
        \item[(3)] $\textbf{(Gauss Formula)}$ For $u \in E(\Omega)$ and $f \in H^1(\Omega)$,
              $$ \left \langle \diverg u,f  \right \rangle + \left \langle u,\nabla f  \right \rangle =\int_{\partial_\Omega}\gamma_v u\cdot\gamma_0f.$$
    \end{itemize}
\end{thm}
\begin{proof}[\textbf{Proof.}]
    For $\phi \in H^{1/2}(\partial\Omega)$, take $w \in H^1(\Omega)$ such that $\phi=\gamma_0w$ (the existence is guaranteed by Theorem \ref{Lift of Trace}). Then for $u \in E(\Omega)$,
    we define an $u-induced$ map from $H^{1/2}(\partial\Omega)$ to $\mathbb{R}$, such that
    $$
        X_u(\phi)=(\diverg u,w)+(u,\text{grad} w).
    $$
    Such $X_u$ is well defined. If $\gamma_0w_1=\gamma_0w_2=\phi$, then $\gamma_0(w_1-w_2)=0$. By Theorem \ref{Kernel of Trace}, $(w_1-w_2)\in H^1_0$.
    Hence there exists $\{p_n\}\subset\cs(\Omega)$, such that $p_n \to (w_1-w_2)$. Since $X_u(p_n)=0$, take $n \to \infty$ and we have $X_u(w_1-x_2)=0$.

    Hence, just take $w=l_{\Omega}\cdot\phi$. Then
    $$|X_u\phi|\leq \|u\|_E\|w\|_{H^1}\leq c_0 \|u\|_E\|\phi\|_{H^{1/2}(\partial \Omega)},$$
    where the first step follows from Schwarz Inequality and the second from the boundedness of $\|l_{\Omega}\|$.

    Therefore, $X_u \in H^{-1/2(\partial\Omega)}$. Then let
    $$\gamma_v:u\mapsto X_u$$.
    It's clear that $\gamma_v$ is a bounded linear map that satisfies $(2)$.

    For $(1)$, take $u \in (\cs(\overline{\Omega}))^3$, $h \in \cs(\overline{\Omega})$.
    Then
    $$
        X_u(\gamma_0h)=\int_{\Omega}\text{div}(wh)=\left \langle u\cdot n,h  \right \rangle|_{\partial\Omega}= \left\langle u \cdot n,\gamma_0 h \right\rangle|_{\partial\Omega},
    $$
    where the last step follows from Stokes Formula.

    Since $\gamma_0(\cs(\overline{\Omega}))$ is a dense subspace of $\gamma_0(H^1(\Omega))=H^{1/2}(\partial\Omega)$, 
    we have $\forall \phi \in H^{1/2}(\partial\Omega)$,
    $X_u(\phi)=\left\langle u\cdot n,\phi \right\rangle|_{\partial\Omega}$, which proves $(1)$. Then combining $(1)$ and $(2)$ yields $(3)$.
\end{proof}

\begin{remark}
    Actually, such $\gamma_v$ is unique because $E(\Omega)$ is the completion of $(\cs(\overline\Omega))^3$.
\end{remark}

\subsection{The Gradient Field and De Rham's Theorem}
\begin{lemma}\label{Ineq of Necas}
    In the space $L^2 (\Omega)$, the $L^2$ norm actually is equivalent to
    the sum of the $H^{-1}$ norm of the function and of its gradient.
    Since the proof is complicated, we only display the conclusion.
    \begin{itemize}
        \item[(1)]     {\textbf{(Nečas Inequality)}}
              Assume $\Omega$ is a Lipschitz domain with compact boundary.\\
              Define the space
              $\chi(\Omega)=\{p\in H^{-1}(\Omega),\nabla p \in (H^{-1}(\Omega))^3\}$, endowed
              with the norm $\|p\|_{\chi}=\|p\|_{H^{-1}}+\|\nabla\|_{H^{-1}}$. Then we have
              $\chi(\Omega)=L^2(\Omega)$ and there's a $C >0$ such that
              $$
                  \|p\|_{L^2(\Omega)}\leq C \|p\|_{\chi_{\Omega}} \text{ , } \forall p \in L^2(\Omega).
              $$
        \item[(2)]  {\textbf{(Nečas-Poincaré Inequality)}}
              Assume $\Omega$ is a connected, bounded, Lipschitz domain with compact boundary. Then there's a $C > 0$ such that
              $$
                  \|p\|_{H^{-1}(\Omega)}\leq C \left(\frac{1}{|\Omega|}\left|\int_{\Omega}pdx\right|+\|\nabla p\|_{H^{-1}(\Omega)}\right) \text{ , } \forall p \in L^2(\Omega).
              $$
    \end{itemize}
\end{lemma}

Applying the Nečas related inequalities, we can make a fundamental
characterization of a gradient ﬁeld, which is called \textbf{de Rham’s theorem}.
Before that there's still some preparation to make.

\begin{thm}\label{Closed of Gradient}
    Assume $\Omega$ be a bounded Lipschitz domain of $\mathbb{R}^3$.
    Then the image of
    the gradient operator
    $$
        \nabla (L^2(\Omega) )=\{f\in H^{-1}(\Omega)^3,\exists p \in L^2(\Omega),f=\nabla p\}
    $$
    is closed in $H^{-1}(\Omega)^3$.
\end{thm}
\begin{proof}[\textbf{Proof.}]
    Take $(\nabla p_n )$ that converges in
    $(H^{-1} (\Omega)) ^3$ with $p_n \in L^2 (\Omega)$.
    Since $\Omega$ is bounded, assume $m(p_n ) = 0$ without changing $\nabla p_n$ .
    Then apply Lemma \ref{Ineq of Necas} and we have
    $$
        \|p_{n+k}-p_n\|_{L^2} \leq C\|\nabla (p_{n+k}-p_n)\|_{H^{-1}}
    $$
    Since $(\nabla p_n )_n$ is a Cauchy sequence in $(H^{-1})^3$, $(p_n)$ is also
    a Cauchy sequence in $L^2 (\Omega)$. Hence there exists a $p \in  L^2 (\Omega)$ as
    a limit of $(p_n)$. Therefore, $\nabla p_n \to \nabla p$ in $(H^{-1} (\Omega))^3$ and
    the claim is proved.
\end{proof}
Now we can prove a weak form of de Rham Theorem.
\begin{thm}\label{Weak de Rham}
    Assume $\Omega$ be a connected, bounded Lipschitz domain of $\mathbb{R}^3$ with compact boundary.
    Let f be in $H^{-1} (\Omega)^3$ such that
    $$
        \left\langle f,\phi \right\rangle_{H^{-1},H^1_0}=0,\forall \phi \in (H^1_0(\Omega))^3 \text{ that div}\phi=0.
    $$
    Then there exists a unique $p \in L^2 (\Omega)$ apart from a constant, such
    that $f = \nabla p$.
\end{thm}
\begin{proof}[\textbf{Proof.}]
    Denote $Y=\nabla( L^2(\Omega))$. Denote $Z=\{\phi \in (H^1_0)^3,\diverg \phi =0\}$
    Then we need to show that if $f \in Z^\bot $, then $f \in Y$. By Theorem \ref{Closed of Gradient},
    $Y$ is a closed subspace. Since $Y$ is also reflexive, $Y=\overline{Y}=Y^{\bot\bot}$.
    Therefore, we only need to show that $Y^\bot \subset Z$.

    Take $u \in Y^\bot \subset (H_0^1 (\Omega))^3$. Then
    $$
        \left\langle \nabla p,u \right\rangle_{H^{-1},H_0^1} = 0, \forall p \in L^2(\Omega).
    $$
    Also, for any $p \in L^2(\Omega)$,
    $\left\langle \nabla p,u \right\rangle_{H^{-1},H_0^1} = -\int_{\Omega} p(\diverg u) dx.$
    Hence $\int_{\Omega} p(\diverg u )dx=0$.

    Therefore, choosing $p = \diverg u$ in this formula exactly yields that $\diverg u = 0$ and thus
    $u \in Z$.
\end{proof}

Now we can now prove a stronger version with weaker assumptions. Indeed, if we further assume that
$f$ cancels against $\cdf(\Omega)$, then it will still give the same result.

\begin{thm}\label{Strong de Rham}{\textbf{(de Rham's Theorem)}}
    Assume $\Omega$ be a connected, bounded Lipschitz domain of $\mathbb{R}^3$ with compact boundary.
    Let f be in $H^{-1} (\Omega)^3$ such that
    $$
        \left\langle f,\phi \right\rangle_{H^{-1},H^1_0}=0,\forall \phi \in \cdf(\Omega).
    $$
    Then there exists a unique $p \in L^2 (\Omega)$ apart from a constant, such
    that $f = \nabla p$.
\end{thm}
\begin{proof}[\textbf{Proof.}]
    Take $(\Omega_n)$ be an increasing sequence of regular and connected open sets
    included in $\Omega$ and covering $\Omega$. For all divergence free $u_n \in (H^1_0(\Omega))^3$,
    make a smoother $u_{n,\epsilon}=u_n\star\eta_\epsilon$ for them and thus $u_{n,\epsilon}\in\cdf$.
    Then
    $$
        \left\langle f,u_n \right\rangle_{H^{-1},H^1_0}=\lim_{\epsilon \to 0}\left\langle f,u_{n,\epsilon} \right\rangle_{H^{-1},H^1_0}=0,
    $$
    which yields the restriction of $f$ onto $\Omega_n$ is the gradient of some $p_n \in L^2(\Omega_n)$ by Theorem \ref{Weak de Rham}.
    Furthermore, for all $n$, the restriction of $p_{n+1}$ onto $\Omega_n$ has the same gradient as $p_n$
    because of the uniqueness. Therefore we can choose $(p_n)$ so that $p_{n+1}-p_n$ is always zero in $\Omega_n$.
    Hence such $(p_n)$ define a suitable $p\in L^2_{loc}(\Omega)$ that $f=\nabla p$. Then it just remains to show such $p$ indeed belongs to $L^2(\Omega)$, whose solution is to cover $\Omega$ with a 
    finite number of star-shaped open sets and to repeat the above actions again.
\end{proof}

\section{The Helmholtz-Weyl decomposition of $L^2(\Omega) \subset \mathbb{R}^3$}
\subsection{The Orthogonal Decomposition}
\begin{defn}{\;}
    \begin{itemize}
        \item [(1)] $G = \{u \in (L^2(\Omega))^3,u=\nabla p, p \in H^1(\Omega)\}$
        \item [(2)] $\mathcal{H}=\{u\in(L^2(\Omega))^3,\diverg u=0,\gamma_vu=0\}$
    \end{itemize}
\end{defn}
\begin{thm}\label{Helholtz}
    Assume $\Omega$ be a connected, bounded, Lipschitz domain of $\mathbb{R}^3$.
    Then we have
        $$H=\mathcal{H},\quad H^{\bot}=G.$$
\end{thm}
\begin{proof}[\textbf{Proof.}]
 First we show $H^{\bot}=G$. 
 It's trivial to show $G \subset H^\bot$, since $\forall u \in G,v \in \cdf(\Omega)$, 
 $$
 \left\langle u,v \right \rangle_{L^2}=\int_{\Omega}\nabla p\cdot v dx = -\int_{\Omega}p(\diverg v)=0.
 $$
 Then it suffices to show $H^\bot \subset G$. For $u \in H^{\bot}$, $\forall v \in \cdf$, 
 $$ 
 \left\langle u,v \right \rangle_{L^2}=\left\langle u,v \right \rangle_{H^{-1},H^1_0}=0.
 $$
 Applying Theorem \ref{Strong de Rham}, there exists a unique $p \in L^2 (\Omega)$ apart from a constant, such
 that $f = \nabla p$. Since $u \in (L^2(\Omega))^3$, $p \in H^1(\Omega)$ and thus $u \in G$. 

 Next we show $H=\mathcal{H}$. It's clear that $H \subset \mathcal{H}$, since for any $u \in H$, div$u=0$ and 
 $\gamma_vu=0$ from the continity of $\gamma_v$. 
 
 Then it suffices to show $\mathcal{H} \subset H$. 
 Denote the orthogonal complement of $H$ into $\mathcal{H}$ as $H_1$. 
 Suppose $u \in H_1 \subset H^\bot$. Then applying $G=H^\bot$ and there exists a 
 $p$ in $H^1(\Omega)$ such that $\nabla p = u$. Since $u \in \mathcal{H}$, we have 
 \begin{align*}
    \left\{\begin{array}{lc}\nabla p=\text{div}u=0,&in\;\Omega,\\\frac{\partial p}{\partial v}=\gamma_v(u)=0,&on\;\partial\Omega.\end{array}\right. 
 \end{align*}
 Applying Gauss Formula in Theorem \ref{Existence of normal component} and we have 
 $$
 \int_{\Omega}|\nabla p|^2dx=0,
 $$
 and thus $u=\nabla p =0$. Hence $H_1=\{0\}$ and we conclude that $H=\mathcal{H}$.
\end{proof}
\begin{thm}\textbf{(The Helmholtz-Weyl Decomposition)}
    The decomposition
$$
    (L^2 (\Omega))^3 = H \oplus  G,
$$
    is called the Helmholtz-Weyl decomposition of the space $(L^2(\Omega))^3$. 
    The orthogonal projection from  $(L^2(\Omega))^3$ onto $H$ is known as the 
    Leray projection and denoted as $\mathcal{P}$.
\end{thm}













\end{document}
